Quote from: telecom on April 12, 2018, 10:23:06 PM
Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?

Just have in mind that the potential energy is not more than the energy you put in to "create" this potential energy.
If you throw a ball upwards, its mass acceleration while the ball is still in the hand, require a given amount of energy to achieve a given velocity.
This velocity is the reason why the ball reach a given hight. The mass is the same all the way, and the velocity of the ball when it comes back and hits your hand, is the same, or less due to air resistance.
So no gain in potential energy.


Vidar

Yes, but there is a gain in kinetic energy which is transferred into potential.
How are you going to harvest kinetic energy?

Telecom's question: How are you going to harvest kinetic energy?

I am going to let the technology used in hydroelectric plants harvest the kinetic energy.

Lets throw a 147 gram ball up into the air 20 meters. That will require a velocity of (d = ½ v²/a) 19.81 m/sec. This is (1/2mv²) 28.83 joules.

We will wrap a 20 meter string around a (147 gram *39) 5733 gram rim mass wheel. We will attach the ball to the string and drop the ball 20 meters.

The ball and rim will accelerate at (9.81m/sec/sec / 40)  .24525 m/sec/sec.

The ball and rim will have a final velocity (after the ball has dropped 20 meters) of (d = ½ v²/a) 3.13 m/sec.

This is 5880 grams moving 3.13 m/sec. This is 18.4 units of momentum.

We will now transfer this 18.4 units of momentum to the 147 gram ball by using the cylinder and spheres.

The ball will now be moving 125 m/sec.

This is (1/2mv²) 1151 joules of energy; and you started with 28.8 joules.

And it is better to stack the balls and drop the stack of 20 balls.

The mass of 147 grams is about that of a baseball; so this can be, and is, applied to a real experiment. 


At Sault St. Marie the Saint Marys River drops 2000 metric ton of water 7 meters every second. This is 30 MW

So we will increase the drop to double the drop (50 meters) at Niagara: 100 meters.

You could drop 2000 metric tons of mass with a head of 200,000 metric tons every second.

Less than 3% of the energy produced would be needed to reload the system and bring one 2000 metric ton unit back to the 100 meter top. This would generate at least 500 MW.

There a three formulas that describe motion. Modern physics claims that all three formulas are conserved quantities. It is claimed that all three are simultaneously conserved. A mathematical evaluation of the three formulas would quickly reveal that they cannot all be conserved: the evaluation would show that only one of the three formulas could be conserved.

The formulas are mv: 1/2mv²: and mvr. These are linear Newtonian momentum: Kinetic Energy: and Angular Momentum.

So in a closed system where we have motion interactions all three formulas would allegedly remain the same.
 
First let's take the ballistic pendulum experiments. A 1 kilogram mass moving 50 m/sec collides with a 19 kilogram mass at rest. The resultant motion is a 20 kilogram mass moving 2.5 m/sec. The mv is conserved. The 1/2mv² loses 95% of the physical motion to unrecoverable heat. Now heat is considered motion but a return to the original condition of 1 kilogram moving 50 m/sec would expose this heat content to be a myth. There is no mechanism for recovery of the heat. The cylinder and spheres proves that the yo-yo despin can be returned to its original state; of the slowly spinning satellite.  This totally eliminates 1/2mv² as a conserved quantity. A number is rarely equal to its square.

The mvr can be exposed by interrupting the string of a rotating mass on the end of a string. Rotate a soft ball on the end of a string, have someone interrupt the string somewhere down its length. As the ball begins rotating in the smaller circle angular momentum is decreased; linear Newtonian momentum remains the same.  Mutilating a conserved number by different r's will not give you the same conserved quantity.
 
Of the three laws only one is conserved; mv.

https://www.youtube.com/watch?v=oeG7RcSodn8

The mass difference, of cylinder to sphere, is only about 3 to 1.There are not two spheres only one. Half the motion is on the other side of the center of mass still in the mass of the cylinder. The motion is complex.

All the motion comes from gravitational potential energy. The sphere's tether gets very long, but the sphere is moving very fast.

If angular momentum is conserved the arc velocity of the sphere would become only .375 of the original speed of the rotating cylinder; because of the long radius. It is obvious that the sphere velocity is much higher. And the gravitationally source is not at the center of rotation: as is the situation in space.

For kinetic energy to be conserved the top velocity of the sphere would have an increase from only 1 to 1.73.

If momentum were to be conserved the velocity of the sphere would have an increase from 1 to 3. These higher speeds seem more apparent; especially since the sphere can stop and lift a falling cylinder.

One might ask why the sphere does not return the cylinder to the top starting position. Well the cylinder's spinning first has to be stopped and that takes time. It takes time for the sphere to restart the spin of the cylinder in the opposite direction. And it would take time to bring the cylinder back up to another stopped position at the top. Going from stop to stop would take a ton of time. And in all of this time the cylinder is under gravitational acceleration.  It is amazing that the single sphere gets the cylinder as far back up as it does. 

Sorry: that the release position is not in view; it starts just above the viewing area. There are two strings on the cylinder and the sphere is in the center.